# What exactly is an ideal hexagonal close-packed structure?

The ideal hexagonal close-packed structure was said to have a fixed $$c/a$$ ratio. However, in terms of three-dimensional Bravais lattices, it looked like Primitive (P) Hexagonal, and there didn't seem to be a reason for which $$c$$ could not be stretched much longer.

What exactly is an ideal hexagonal close-packed structure? Why can't it be stretched longer in $$\vec{c}$$ direction? Also, why does it has a packing fraction of $$\frac{\sqrt{2}}{6}\pi$$?

Close packing refers to the arrangement of spheres in a 3-dimensional space. The close-packed structure is obtained by maximizing the average density of "spheres" with a fixed radius. This arrangement have an hexagonal structure, and the simplest realizations are either FCC or HCP lattices. The packing fraction $$\approx0.74$$ is the fraction of volume filled by the spheres, and the whole volume occupied by the structure (which includes both filled and empty space).

If you stretch the $$c$$ axis you change the density of spheres, i.e. the number of spheres (atoms) per unit volume. In particular if the ratio $$c/a$$ is longer than the optimal value $$c/a=\sqrt{8/3}$$, your packing fraction $$f$$ will be smaller than the $$\sqrt{2}\pi/6$$.

The reason why the close-packed structure has packing fraction equal to $$\sqrt{2}\pi/6$$ can be obtained by simple geometry (see below). However to demonstrate that this is the ideal ratio which maximize density is a highly nontrivial mathematical problem which took hundreds of years to be solved.

Let's calculate the packing fraction. Please refer to the figure below. Let's consider the FCC (face-centered cubic) structure. In the unit cell, there are six spheres (or atoms) intersecting the cube surfaces, but only with one half of their volume is inside the unit cell: in total 3 whole sphere volumes. There are eight other spheres at the cube corners, but only with one eighth of their volume inside the unit cell: in total 1 sphere volume. All taken together, in total there are four spheres in the unit cell. So the total volume filled by the spheres is $$V_S=4 \cdot \frac{4}{3} \pi r^3$$ where $$r$$ is the sphere radius. Now, the diagonal of the unit cell is equal to $$\sqrt2 a$$ where $$a$$ is the length of one one side the unit cell (the unit cell is cubic). The diagonal accommodates 4 sphere radii, hence $$r=\frac{\sqrt{2}}{4} a$$. The packing fraction is the ratio between the volume $$V_S$$ occupied by the spheres and the total volume of the unit cell. The total volume of the unit cell in terms of the radius is $$V=a^3=\left(\frac4{\sqrt2}\right)^3r^3$$ Finally one can obtain the packing fraction $$f=\frac{V_S}{V}=\frac{\pi\sqrt2}6$$

Another thing to say: The close-packed structures FCC (face-centered cubic) and HCP (hexagonal close-packed) differ in the arrangement of stacked planes along the $$c$$ axis. FCC is a Bravais lattice, but HCP is not a pure Bravais lattice, because there are two inequivalent lattice positions.  • Only a remark. HCP is not a Bravais lattice, but FCC is indeed one. Sep 7, 2020 at 22:58
• @ClaudioSaspinski Thank you for the remark. I edited the answer. Sep 8, 2020 at 7:05

Indeed, the c-axis of an hexagonal cell can be easily stretched. For hard spheres, the ratio would be $$\sqrt{8/3} \approx 1.866$$, with the same distance between close-packed planes as in cubic close packed crystals. The figure on the left shows what that that looks like.

But metals are more complicated than hard spheres. There are not only nearest-neighbor interactions. That is how in zinc and cadmium that ratio can be larger, which I tried to visualize in the figure on the right. It can also be smaller, which would be difficult to make a figure of.

This does not change the symmetry of the lattice. • if you stretch the $c$ axis, the packing fraction becomes larger, than by definition you do not have a closed packing structure Sep 8, 2020 at 6:57

Look at this picture:

A nice visualization of a hexagonal Braivais lattice. When you pack them together the height $$c$$ is of importance.
But, you're asking about an ideal packed case. I guess in this case, the parallelograms must obey: $$a=b=c$$.

If we take the volume of the 3-d parallelogram, expresses in $$a$$, $$b$$, and $$c$$, which are al equal. So the volume will be $${(a\sin{60}})^3$$. The volume of the whole hexagonal form will be $$2.5{(c\sin{60})}^3$$.
You have to subtract the volume of the parallel from the volume of the hexagonal, which gives the packing fraction in for an ideal hexagonal:

$$2.5{(c\sin{60})}^3 - {(c\sin{60}})^3=0.675c-0.027c=0.648c$$

Obviously and logically this doesn't equal $$\frac{\sqrt{2}}{6}\pi$$, although it's close. It's smaller because no space exists between the hexagonals.

• The packing fraction do not depend on a and b. Also, your last equation is dimensionally wrong. Sep 7, 2020 at 8:53
• Who is that guy in the picture? Is he relevant for the answer? Sep 7, 2020 at 8:54
• The equation is still wrong. The packing fraction cannot depend on the lattice constants. Sep 7, 2020 at 9:18
• The hexagonal Bravais lattice and hexagonal close packing are not the same. Sep 7, 2020 at 14:59
• HCP is a the hexagonal Bravais lattice with a two atom basis, resulting in en.m.wikipedia.org/wiki/Atomic_packing_factor, a structure like stacking cannonballs (hence close packing). The pure hexagonal lattice with solid spheres does not pack as densely - they can shift over to nestle closer to the layer below (making it hcp or fcc in the process). Sep 7, 2020 at 15:08